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start by finding y'
y'=6x^2+8x-5
Yes, find the slope at x = 1
@ganeshie8 after you help this person with her question can you please help me with mine
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sorry i suck at english :|
8?
y'(x) =6x^2+8x-5 slope at x = 1 y'(1) = 6(1)^2 + 8(1) - 5 = 6 + 8 -5 = 9
so what is the y intercept then
so slope, m = 9 u knw a point (1, -2) u can write teh equation of tangent in point slope form
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y-1=9(x--2) y-1=9(x+2)
nope, do it in ur regular way may be
y = mx + b y = 9x + b
solve b, by putting (1, -2) above
-2=9*1+b -2=9+b -11=b
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Yes, so the equation of tangent line at (1, -2) is :- y = 9x - 11
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